INTEGRATION BY PARTS (IBP)
(subtitle: the ANTIPRODUCT RULE)
The BASIC TOOL (from our text Hoffmann)

f(x) g'(x) dx = f(x)^{.}g(x)  

f '(x) g(x) dx 


Prof. M^{c}Farland prefers an equivalent statement found in many texts.
The following popular form of our BASIC TOOL is easier to remember:
The BASIC TOOL: alternate form 

u v' dx = u^{. }v  

u ' v dx 


If one differentiates both sides of either equation above, the result is
the PRODUCT RULE for derivatives, hence the name "ANTIPRODUCT RULE".
The understanding in both forms is that "x" is the basic variable of
all functions:
f(x), g(x), and also u and v.
WHEN DO I USE INTEGRATION BY PARTS ?
Try INTEGRATION BY PARTS when all other methods have failed:
"other methods" include POWER RULE, SUM RULE, CONSTANT MULTIPLE RULE, and SUBSTITUTION.
For this method to succeed, the integrand (between _{} and "dx") must be a product of two quantities:
you must be able to differentiate one, and antidifferentiate the other.
Also, after applying the BASIC TOOL above, the
new integral you obtain should be easier to evaluate than the original.
If any of these requirements is not met, abandon your choice of u and
v' and be prepared even to abandon this method. 
Example 1. (Hoffmann, Pg 425, #6) In the following presentation, the
choices for u and v' must be made at the start. These are colorcoded,
and the BASIC TOOL (copied from above) is positioned nearby for comparison.
Notice that isolated differentials are not used in this solution.
Example 2.
Compare this solution with a different solution
of the same problem. Especially, note that the 2 differing methods appear
to produce different final answers. However, it is possible to show the
two answers are indeed equal, by algebraically altering their forms.

x^{.}(x+1)^{8}dx 

= 
x^{.}(1/9)(x+1)^{9} 
 

1^{.}(1/9)(x+1)^{9}dx 

= 
(1/9)x(x + 1)^{9}  (1/9)(1/10)(x+1)^{10} + C 



u ^{.} v' dx 

= 
u ^{.} v 
 

u' ^{.} v dx 

All work is colorcoded so equal
numbers have the same color 


LET u = x 
THUS u' = 1 
THUS v = 

v' = 

(x+1)^{8}dx = (1/9)(x + 1)^{9} 

and let v' = (x+1)^{8} 




This page last updated 14 March 2001 
